I would like to show that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as

$$ h(t) = \left\{ \begin{array}{ll} 0 & \mbox{if } t \leq 0 \\ \exp{(-\frac{1}{t})} & \mbox{if } t > 0 \end{array} \right. $$

is $C^{\infty}$. It seems to me that the problem is at $t = 0$ and I assume I have to use the definition of the derivative and induction, but I don't know how to generalize this result to $\infty$.


By induction show that for $t>0$ we have that $h^{(n)}(t)$ is a polynomial in $\frac1t$ times $\exp(-1/t)$ (and of course $h^{(n)}(t)=0$ for $t<0$). Conclude from this fact (for $n-1$) that $h^{(n)}(0)=0$ and $h^{(n)}$ is continuous. In other words, show that there exists polynomials $f_n(X)\in\mathbb R[X]$, $n\in\mathbb N_0$ such that $$ h^{(n)}(t)=\begin{cases}0&\text{if }t\le 0\\f_n(\tfrac 1t)\exp(-\tfrac 1t)&\text{if }t>0.\end{cases}$$

  • $\begingroup$ I was thinking on something like that, but is it enough to show that $h \in C^{\infty}$. $\endgroup$ – Tim Brenen Nov 13 '13 at 12:27

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